Whoops, sorry, was thinking charge on surface of sphere.
If uniformly distributed throughout the volume, then my answer is the same outside the sphere
However inside the sphere you will surround chagres the will look as if they are at the center
The amount you surround will be the volume at radius r times the charge density.
V = (4/3) pi r^3
q =V (charge density)= (4/3) pi r^3 (p) (r^2/a^2)
which is
(4/2) p *pi ^5/a^2
that times is q
E = 9*10^9 q/r^2
= 9 * 10^9 (4/3)p* pi r^3/a^2

when you reach r = a
you get
E = 9*10^9 (4/3) p* pi r^3 /a^2
which is of course
9*10^9 ( q total)/r^2 from then on

The idea is that inside the sphere, the only charge that counts is what is inside the radius you are at. That looks like it is at the center and is the charge density times the volume inside the radius you are at.

Outside the sphere, the entire charge, the density times the entire volume of the sphere (4/3) pr a^3 * density is inside your radius and no longer increases/